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I don't know how to solve this limit. What should I do?

$$ \lim_{x\to 0} {\sqrt{x^3+2x+1}-\sqrt{x^2-3x+1} \over \sqrt{4x^2-3x+1} - \sqrt{2x^3+6x^2+5x+1}} $$

Thank you!!

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  • $\begingroup$ You can start by rationalizing the denominators. $\endgroup$
    – David H
    Aug 24, 2014 at 5:02

4 Answers 4

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Multiply the given expression by a special kind of $1$ :

$1 = \frac{\sqrt{x^3+2x+1}+\sqrt{x^2-3x+1}}{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}} . \frac{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}{ \sqrt{x^3+2x+1}+\sqrt{x^2-3x+1} }$


$\lim \limits_{x\to 0} \frac{\sqrt{x^3+2x+1}-\sqrt{x^2-3x+1}}{ \sqrt{4x^2-3x+1} - \sqrt{2x^3+6x^2+5x+1}} \\= \lim \limits_{x\to 0} {\sqrt{x^3+2x+1}-\sqrt{x^2-3x+1} \over \sqrt{4x^2-3x+1} - \sqrt{2x^3+6x^2+5x+1}} .\frac{\sqrt{x^3+2x+1}+\sqrt{x^2-3x+1}}{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}} .\frac{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}{ \sqrt{x^3+2x+1}+\sqrt{x^2-3x+1} } \\= \lim \limits_{x\to 0} \frac{x^3-x^2+5x}{-2x^3-2x^2-8x} .\frac{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}{ \sqrt{x^3+2x+1}+\sqrt{x^2-3x+1} } \\= \lim \limits_{x\to 0} \frac{x^2-x+5}{-2x^2-2x-8} .\frac{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}{ \sqrt{x^3+2x+1}+\sqrt{x^2-3x+1} } $

plugin x = 0

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    $\begingroup$ It is not affecting the answer of this problem but still want to point out that finally the denominator has $ -2x^2-2x-8 $ as one of the expressions $\endgroup$
    – Vikram
    Aug 24, 2014 at 6:00
  • $\begingroup$ thanks for catching the mistake :) will correct it now $\endgroup$
    – AgentS
    Aug 24, 2014 at 8:19
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As David H commented, rationalizing the denominator would be a good starting point.

Now, if you know Taylor series, the problem starts to be simple since, around $x=0$, you have $$\sqrt{x^3+2x+1}=1+x+O\left(x^2\right)$$ $$\sqrt{x^2-3x+1}=1-\frac{3 x}{2}+O\left(x^2\right)$$ $$\sqrt{4x^2-3x+1}=1-\frac{3 x}{2}+O\left(x^2\right)$$ $$\sqrt{2x^3+6x^2+5x+1}=1+\frac{5 x}{2}+O\left(x^2\right)$$ So, numerator is $$ \frac{5 x}{2}+O\left(x^2\right)$$and denominator is $$-4 x+O\left(x^2\right)$$

I am sure that you can take from here.

All of the above has been done using the fact that, if $x$ is small $$\sqrt{a+bx+cx^2+dx^3+ex^4+\cdots}\simeq \sqrt{a+bx}=\sqrt{a}+\frac{b x}{2 \sqrt{a}}+O\left(x^2\right)$$

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    $\begingroup$ Kind of a tangent, but do you know if there are any situations in which a Taylor series gives you the wrong answer? $\endgroup$
    – user541686
    Aug 24, 2014 at 9:00
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    $\begingroup$ I don't know any but I know so little .. ! Cheers $\endgroup$ Aug 24, 2014 at 9:01
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Well, for small numbers, $$ {x^3 << x^2 << x << 1} $$ So the given integral basically reduces to, $$ \lim_{x\to 0} {\sqrt{2x+1}-\sqrt{1-3x} \over \sqrt{1-3x} - \sqrt{5x+1}} $$ as higher powers become negligible. This is the same as ignoring lower exponents of $x$ for limits that tend to infinity.

Recalling that for $$ { x << 1} $$ we have, $$ {(1 + x)^{1/2} = 1 + x/2 - x^2 / 8 + ...} $$ We can similarly ignore higher exponents of x, thus we have, $$ {(1 + x)^{1/2} \approx 1 + x/2} $$ Thus the limit simplifies to, $$ \lim_{x\to 0} {(1+x) - (1- 3x/ 2 )\over (1- 3x/2) - (1 + 5x/2)} $$ Which reduces to $-5 \over 8$$=-0.625$.

Trial value, putting $x = 0.1$, we have $-0.65017361$ for $x = 0.01$, we have $-0.62692914$, as we can see the limit converges to the required value.

Hope this helped :)

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The expression is of the form $\frac{\sqrt a-\sqrt b}{\sqrt c-\sqrt d}$

Rationalise/simplify as $\sqrt a-\sqrt b\times \frac{\sqrt a+\sqrt b}{\sqrt a+\sqrt b}\times \frac{1}{\sqrt c-\sqrt d}\times \frac{\sqrt c+\sqrt d}{\sqrt c+\sqrt d}$

Answer is $\frac{-5}{8}$

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